The work done by the net force on a particle equals the change in its kinetic energy: W_net = ΔK = K_f - K_i = ½ m (v² - v₀²). Holds for both constant and variable forces and for one or more dimensions.
-- NCERT Class 11 Physics, Ch. 5, p. 73Work Energy Theorem
8 MCQs2 revision cards9-step worked example
Source: NCERT Work, Energy and PowerOfficial key: NTA-verifiedLast updated: 20 Sep 2026
Work Energy Theorem, explained for NEET
The word that decides every work-energy question is net. The theorem reads W_net = ΔK, and W_net means the vector-sum work of every force acting — not gravity alone, not the applied force alone. A common failure is computing the work of one convenient force and setting it equal to ΔK. If three forces act and you only account for two, the equation is simply not the theorem.
NCERT Class 11 Physics Chapter 5 states it on page 73: the change in kinetic energy of a particle equals the net work done on it, W_net = ½m(v² − v₀²).
Two features make it worth more than the kinematic equations it replaces.
First, it is indifferent to the path and the time taken. You never need the acceleration, never need how long the motion lasted, never need whether the force was steady. Only the start speed, the end speed, and the total work matter.
Second, it holds for variable forces, which v² = u² + 2as does not. That is why a "find the speed after the force changed partway" question is a work-energy question, not a kinematics question.
A related habit worth breaking: forces perpendicular to the displacement contribute nothing to W_net. The normal reaction on a block sliding along a level floor does zero work, since cos 90° = 0. Students often add N·s into the tally and get a wrong ΔK. Include the force in your force list, then let the angle zero out its contribution — don't drop the force, and don't award it work it never did.
Sign is the other half. Negative net work means the body slows: ΔK is negative, so v < v₀. A body dragged to rest by friction has W_net = −K_initial, not +K_initial.
Watch out for the one-force substitution. Before you write W_net, list every force acting, then compute each one's work — including the zeros.
Can you answer these Work Energy Theorem MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The work-energy theorem, as stated in NCERT Class 11 Physics Chapter 5, equates the change in kinetic energy of a particle to:
Show answer and why every option is right or wrong
Answer: C. The statement on page 73 of NCERT Class 11 Physics Chapter 5 specifies the net work — the vector-sum work of every force acting. C is correct.
Why A is wrong: A is wrong because the applied force is only one contributor; friction, gravity and normal reaction all enter the tally when they do work.
Why B is wrong: B is wrong because the theorem places no restriction on force type — non-conservative work counts in W_net exactly as conservative work does.
Why D is wrong: D is wrong because gravity is a single force; using its work alone is the one-force substitution the theorem forbids.
MCQ 2Easy RecallPractice
For which of the following is the work-energy theorem valid, according to the conditions of use given in NCERT Class 11 Physics Chapter 5?
Show answer and why every option is right or wrong
Answer: B. The theorem holds for constant and variable forces alike and in any number of dimensions, as stated on page 73 of NCERT Class 11 Physics Chapter 5. B is correct.
Why A is wrong: A is wrong because it imposes two restrictions the theorem does not carry — the theorem is neither limited to constant forces nor to one dimension.
Why C is wrong: C is wrong because restricting to conservative forces describes mechanical-energy conservation, not the work-energy theorem.
Why D is wrong: D is wrong because variable forces are covered; this is precisely why the theorem outruns v² = u² + 2as.
MCQ 3Easy RecallPractice
In the expression W_net = ½m(v² − v₀²), the symbols v and v₀ denote:
Show answer and why every option is right or wrong
Answer: A. v is the final speed and v₀ the initial speed, so the bracket is (final² − initial²) — the change in kinetic energy per unit ½m. A is correct.
Why B is wrong: B is wrong because reversing the order flips the sign of ΔK, turning a speeding-up body into a slowing-down one.
Why C is wrong: C is wrong because the theorem compares the two endpoints of the interval, not extremes reached anywhere along the way.
Why D is wrong: D is wrong because kinetic energy is built from instantaneous speed at the endpoints, not from averages over stretches of path.
MCQ 4Direct ApplicationPractice
A particle of mass 2.0 kg has its speed changed from 3.0 m/s to 5.0 m/s by the forces acting on it. The net work done on the particle is:
Show answer and why every option is right or wrong
Answer: C. W_net = ½m(v² − v₀²) = ½ × 2.0 × (25 − 9) = 16 J, following page 73 of NCERT Class 11 Physics Chapter 5. C is correct.
Why A is wrong: A is wrong because it computes ½m(v − v₀)² = ½ × 2.0 × 4, squaring the speed difference instead of differencing the squared speeds.
Why B is wrong: B is wrong because it uses the final kinetic energy ½ × 2.0 × 25 = 25 J incorrectly scaled, or omits the ½ from ½m(v² − v₀²) = 2.0 × 16; either way it ignores that only the change counts.
Why D is wrong: D is wrong because it uses m(v − v₀)·something linear in speed; work depends on the difference of squared speeds, not on the speed difference.
MCQ 5Direct ApplicationPractice
A body of mass 4.0 kg moving at 6.0 m/s is brought to rest by the forces acting on it. The net work done on the body during this process is:
Show answer and why every option is right or wrong
Answer: B. W_net = ½ × 4.0 × (0 − 36) = −72 J. The net work is negative because the kinetic energy decreases, per page 73 of NCERT Class 11 Physics Chapter 5. B is correct.
Why A is wrong: A is wrong because it reports the magnitude with a positive sign; positive net work would speed the body up, not stop it.
Why C is wrong: C is wrong because it drops the square on the speed, using ½ × 4.0 × 6.0 × 2 or similar rather than ½m v₀².
Why D is wrong: D is wrong because the kinetic energy genuinely changes from 72 J to zero, so W_net cannot be zero.
MCQ 6Direct ApplicationPractice
A block of mass 5.0 kg slides along a horizontal floor. Three forces act on it: the applied horizontal push, the normal reaction from the floor, and the weight of the block. Which of these contribute to the net work over a horizontal displacement?
Show answer and why every option is right or wrong
Answer: D. Both the normal reaction and the weight are vertical, while the displacement is horizontal, so each makes θ = 90° and W = Fs cos 90° = 0. Only the horizontal push does work. D is correct.
Why A is wrong: A is wrong because two of the three forces are perpendicular to the displacement, and a perpendicular force does zero work however large it is.
Why B is wrong: B is wrong because the weight is vertical here, so it is one of the zero-work forces — not a contributor.
Why C is wrong: C is wrong because it reports W = N·s for the normal reaction on a level surface, which is the classic perpendicular-force error; cos 90° = 0.
MCQ 7Concept TrapPractice
A body is dragged across a rough surface by a horizontal force, and over the journey the body's speed is unchanged. A student concludes that no force did any work. The flaw in this reasoning is that:
Show answer and why every option is right or wrong
Answer: B. W_net = ΔK = 0 constrains the sum, not each term: the applied force does positive work and friction an equal negative amount. B is correct.
Why A is wrong: A is wrong because the theorem places no restriction on force type; friction's work simply enters W_net with a negative sign.
Why C is wrong: C is wrong because constant speed over a real journey means a non-zero displacement, not a zero one.
Why D is wrong: D is wrong because kinetic energy ½mv² is perfectly well defined at any speed, constant or changing.
MCQ 8CalculationPractice
A particle of mass 3.0 kg moving at 4.0 m/s is acted on by two forces along its line of motion: one does +30 J of work and the other does −6.0 J of work over the same displacement. The final speed of the particle is:
Show answer and why every option is right or wrong
Answer: A. A is correct. W_net = 30 − 6.0 = 24 J. The theorem gives ½ × 3.0 × (v² − 4.0²) = 24, so 1.5(v² − 16) = 24, v² = 32 and v = 5.7 m/s to two significant figures. Equivalently the kinetic energy rises from 24 J to 48 J. NCERT Class 11 Physics Chapter 5, page 73.
Why B is wrong: B is wrong because it uses only the +30 J and drops the −6.0 J: 1.5(v² − 16) = 30 gives v² = 36 and v = 6.0 m/s. The theorem needs the NET work, which is the one thing it is for.
Why C is wrong: C is wrong because it drops the mass, solving ½(v² − 16) = 24 instead of ½ × 3.0 × (v² − 16) = 24. That gives v² = 64 and v = 8.0 m/s — the right method with m left out.
Why D is wrong: D is wrong because it uses only the −6.0 J: 1.5(v² − 16) = −6.0 gives v² = 12 and v = 3.5 m/s. Picking one force and ignoring the other is the same error as B, in the other direction.
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Work Energy Theorem: quick recall before you leave
How do you solve a Work Energy Theorem question? A worked example
- 1
Given
• Mass m = 2.0 kg• Initial speed v₀ = 3.0 m/s• Final speed v = 7.0 m/s• Work done by friction W_friction = −18 J
- 2
Required
The work done by gravity, W_gravity, along the chute.
- 3
Concept
The work-energy theorem relates the net work to the change in kinetic energy. When all but one force's work is known, the theorem becomes a way of solving for the missing term. Only gravity, friction and the normal reaction act; the normal reaction is perpendicular to the chute surface at every point, so it does zero work.
- 4
Formula
W_net = ΔK = ½m(v² − v₀²), with W_net = W_gravity + W_friction + W_normal.
- 5
Substitution
½ × 2.0 × (7.0² − 3.0²) = W_gravity + (−18) + 0
- 6
Calculation
ΔK = ½ × 2.0 × (49 − 9) = 1.0 × 40 = 40 J.
So 40 = W_gravity − 18, giving W_gravity = 58 J.
The ½ in ½mv² is a counting factor from the definition of kinetic energy, and the exponent 2 is likewise exact — neither contributes to the significant-figure count. The given quantities carry two significant figures each, so the answer is quoted to two. - 7
Final answer
W_gravity = 58 J (two significant figures).
- 8
Common trap
The trap here is answering 40 J — treating the change in kinetic energy as the work done by gravity. That is the one-force substitution: ΔK equals the net work, and the net work is the sum of three contributions, only one of which is gravity's. A second, subtler version is dropping the normal reaction from the force list entirely instead of listing it and letting its perpendicularity zero it out; on a curved chute, forgetting that N is perpendicular at every point is a real error rather than a bookkeeping nicety.
- 9
Similar NEET-style question
A crate of mass 5.0 kg is pulled along a rough horizontal floor and its speed rises from 2.0 m/s to 4.0 m/s. If friction does −24 J of work over the displacement, find the work done by the applied pull. *(Answer: ΔK = 30 J, so W_pull = 30 + 24 = 54 J.)*
What to remember before solving Work Energy Theorem questions
Which Work Energy Theorem formulas do you need for NEET?
1 formula — click to collapse
Work-energy theorem
The net work done by all forces on a particle equals the change in its kinetic energy. Holds for both constant and variable forces, in 1D and higher dimensions.
| Symbol | Quantity | SI Unit |
|---|---|---|
| W_net | Net work done by all forces | J |
| Delta_K | Change in kinetic energy | J |
| m | Mass of particle | kg |
| v, v0 | Final, initial speed | m/s |
Valid when
- W_net is the NET (vector-sum) work, not work of any one force
- Particle (point-mass) idealisation; for extended bodies handle internal energy separately
Where do students lose marks on Work Energy Theorem?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
3 items — click to collapse
Category: Overthinking
Student assumes proportionality of speed to remaining distance under uniform deceleration. In fact, KE drops linearly with distance (v² is the linear quantity, not v): v² = u² - 2as. Speed-vs-distance is a sqrt-curve, not a line.
When it triggers
Question describes a body decelerating through stages with given speed at one stage; asks for distance to stop or speed at another stage.
How to avoid
Always work with v², not v, when uniform deceleration is in play. The work-energy theorem gives the same answer faster: ½ m v² = work done against constant force over distance.
Category: Overthinking
Student computes ideal power and forgets the (1 - loss_fraction) or efficiency multiplier.
When it triggers
Question gives turbine, motor, or transformer with stated efficiency or loss percentage.
How to avoid
Always read the question for efficiency η or loss%. Useful power P_useful = η × P_input or P_input × (1 - loss). Don't drop the factor even if the rest of the calc is in the unrelated parts of the problem.
Root cause: direction ignored
Correction
Work = F . s = F * s * cos(theta). For a block sliding horizontally, the normal force is perpendicular to displacement (theta = 90 deg, cos = 0), so the work done by N is ZERO. Always check the angle between force and displacement before computing work.
Wrong option pattern
Distractor reports W = N * s for a block sliding on a level surface.
More in Work, Energy and Power: 9 exam traps and mistakes · 9 formulas · 5 question patterns from its other lessons.
Work Energy Theorem questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
How does NEET ask about Work Energy Theorem?
2 recurring patterns from past papers — click to collapse
Two inclines of equal length L and same angle θ (e.g. 45°); one rough (with μ), one smooth. Compare time-of-descent or final velocity at bottom. Smooth: a = g sin θ. Rough: a = g(sin θ - μ cos θ). Use L = ½ a t² or v² = 2aL.
Multi StepMedium
Common distractors
ignores mu cos theta term
Forgetting the friction-along-incline component
Water falls from height h at flow rate dm/dt (kg/s); turbine efficiency or loss factor given; find usable power. Power = dm/dt × g × h × (1 - loss_fraction). Common shape: 60 m, 15 kg/s, 10% loss → P = 15 × 10 × 60 × 0.9 = 8.1 kW.
Multi StepEasy
Common distractors
forgets loss factor
Default to ideal-case formula
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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